Problem involving Coulombs Law; suspended masses

[adkinje]

Two point particles, each of mass m and charge q are suspended from a common point by threads of length L. Each thread makes an angle [tex]\theta[/tex] with the vertical. (I attached a diagram to help).

I must show that
[tex]q=2L\sin\theta\sqrt{\frac{mg}{k}\tan\theta}[/tex]

I start out by writing the force sum in the coordinate system, and then I solve for q:

[tex]0=mg+k\frac{q^2}{R^2}[/tex]

[tex]=mg+k\frac{q^2}{4L^2\sin^2\theta}[/tex]

[tex]q^2=-4L^2\sin^2\theta\frac{mg}{k}[/tex]

I'm confused by the negative sign, if I ignore it I get:

[tex]q=2L\sin\theta\sqrt{\frac{mg}{k}}[/tex]

The [tex]\tan\theta[/tex] is missing in my solution.

 

[Doc Al]

I start out by writing the force sum in the coordinate system, and then I solve for q:

[tex]0=mg+k\frac{q^2}{R^2}[/tex]

Forces are vectors--direction matters. Hint: Consider the horizontal and vertical components of the forces.

Don't forget the tension force exerted by the thread.

 

[adkinje]

Forces are vectors--direction matters. Hint: Consider the horizontal and vertical components of the forces.

Don't forget the tension force exerted by the thread.

lol, DUH the tension! I can't believe I left that one out. I pulled it togther and obtained the given expression. Thanks for your help.